Outline. Classical Control. Lecture 2

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Transcription:

Outline

Outline Outline Review of Material from Lecture 2 New Stuff -

Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New Stuff -

Review of Lecture System Performance Effect of Poles Analyzing System Performance System types Continuous system Discrete system Analysis domain Time-domain specifications Frequency-domain specifications Different periods Dynamic (transient) responses Steady-state responses

Review of Lecture System Performance Effect of Poles Continuous Time - Dynamic Response Impulse Response Step Response.2 0.9 0.8 0.8 0.7 Amplitude 0.6 Amplitude 0.6 0.5 0.4 0.4 0.2 0.3 0.2 0 0. 0.2 0 2 4 6 8 0 2 Time (sec) 0 0 2 3 4 5 6 7 8 9 0 num=[]; den=[ 2 ]; num2=[ 2]; den2=[ 2 3]; impulse(tf(num,den), b,tf(num2,den2), r ) step(tf(num,den), b,tf(num2,den2), r ) Time (sec)

Review of Lecture System Performance Effect of Poles Continuous Time - Dynamic Response Step Response Amplitude.4.2 0.9 0.6 0.4 M p % t p t s Overshoot (M p ) Rise time (t r ) Settling time (t s ) Peak time (t p ) 0. t r 0 0 2 4 Time (sec) 6 8 0 2

Review of Lecture Natural Frequency and Damping System Performance Effect of Poles

Review of Lecture Poles in the Continuous s-plane System Performance Effect of Poles 2.5 0.86 0.76 0.64 0.5 0.34 0.6.6.4.2 0.8 0.5 0.94 0.985 0.6 0.4 0.2 0 0.985 0.2 0.5 0.94 0.4 0.6.5 0.86 0.76 0.64 0.5 0.34 0.6 0.8.2.4.6 2.6.4.2 0.8 0.6 0.4 0.2 0

Review of Lecture Poles in the Continuous s-plane System Performance Effect of Poles

Outline Review of Lecture Review of Material from Lecture System Performance Effect of Poles 2 New Stuff -

Review of Lecture Short description of different reference inputs Position reference Servo-motors Position tracking (robots) Velocity reference AC-motors Vehicles Acceleration reference Space shuttle (during launch) Force exerting systems

Review of Lecture Conversion from Time Domain to Laplace Domain Position Time Domain Linear, x(t) Rotational, θ(t) Laplace Linear, X(s) Rotational, Θ(s)

Review of Lecture Conversion from Time Domain to Laplace Domain Velocity Time Domain Linear, v(t) = ẋ(t) Rotational, ω(t) = θ(t) Laplace Linear, V(s) = sx(s) Rotational, Ω(s) = sθ(s)

Review of Lecture Conversion from Time Domain to Laplace Domain Acceleration Time Domain Linear, a(t) = v(t) = ẍ(t) Rotational, α(t) = ω(t) = θ(t) Laplace Linear, A(s) = sv(s) = s 2 X(s) Rotational, A(s) = sω(s) = s 2 Θ(s)

Review of Lecture Why look at steady-state conditions? Different reference inputs Position Velocity Acceleration Steady-state corresponds to the state a system settles in under constant conditions Constant reference can be interpreted differently Constant position (no velocity or acceleration) Constant velocity (ramp/integration in position, no acceleration) Constant acceleration (ramp/integration in velocity, parabola/integration 2 in position)

Review of Lecture Diagram of Feedback System General System + D(s) H(s) G(s) System + D(s) G(s)

Review of Lecture Continuous Time - Steady-State - Laplace transform Final value theorem Steady-state errors F(s) = 0 f (t)e st dt lim f (t) = lim sf(s) t s 0 s n e ss = lim s 0 s n + K n s k

Review of Lecture Continuous Time - Steady-State - Type 0 - K p = lim s 0 D(s)G(s) Step input Ramp input Parabolic input e ss +K p K v K a Static error K p =constant K v = 0 K a = 0 Error +K p

Review of Lecture Continuous Time - Steady-State - Type - K v = lim s 0 sd(s)g(s) Step input Ramp input Parabolic input e ss +K p K v K a Static error K p = K v =constant K a = 0 Error 0 K v

Review of Lecture Continuous Time - Steady-State - Type 2 - K a = lim s 0 s 2 D(s)G(s) Step input Ramp input Parabolic input e ss +K p K v K a Static error K p = K v = K a =constant Error 0 0 K a

Example 4. Review of Lecture System + k p A τs+ Analysis of Steady-State Error System type is?

Example 4. Review of Lecture System + k p A τs+ Analysis of Steady-State Error ( ) System type is 0 L(s) = kpa τs+ Steady-State Error for Step:

Example 4. Review of Lecture System + k p A τs+ Analysis of Steady-State Error ( ) System type is 0 L(s) = kpa τs+ Steady-State Error for Step: e ss = +lim s 0 kpa τs+ = +k pa

Example 4.2 Review of Lecture System + k p + k I s A τs+ Analysis of Steady-State Error System type is?

Example 4.2 Review of Lecture System + k p + k I s A τs+ Analysis of Steady-State Error ( System type is L(s) = (kp+ k I Steady-State Error for Ramp: s )A τs+ = A(kps+k I) s(τs+) )

Example 4.2 Review of Lecture System + k p + k I s A τs+ Analysis of Steady-State Error ( System type is L(s) = (kp+ k I s )A τs+ = A(kps+k I) s(τs+) Steady-State Error for Ramp: e ss = lim s 0 s A(kps+k I ) s(τs+) ) = Ak I

Review of Lecture Continuous Time - Steady-State Laplace transform Final value theorem Steady-state errors F(s) = 0 f (t)e st dt lim f (t) = lim sf(s) t s 0 e ss = lim s 0 T (s) s k

Example 4.3 Review of Lecture System + k p s(τs+) + k t s

Review of Lecture Example 4.3 - Continued Analysis of Steady-State Error T (s) e ss = lim s 0 ( s k k p = lim s 0 s k s(τs + ) + ( + k t s)k p s(τs + ) + ( + k t s )k p = lim s 0 s k s(τs + ) + ( + k t s)k p s 2 τ + s( + k t k p ) = lim s 0 s k s 2 τ + s( + k t k p ) + k p = 0 k = 0 = + k tk p k p k = )

Review of Lecture Book: Feedback Control Problem 4.6 (a+b) Problem 4.8 (a+b) Problem 4.9